What would be the heat released when an aqueous solution containing `0.5 mol` if `HNO_(3)` is mixed with `0.3` mol of `OH^(-1)` (enthalpy of neutralisation is `-57.1 kJ`)

Options

A. 28.5 kJ
B. 17.1 kJ
C. 45.7 kJ
D. 1.7 kJ

Answer

To find the heat released when mixing an aqueous solution containing 0.5 moles of HNO₃ with 0.3 moles of OH⁻, we can follow these steps:

### Step 1: Understand the Reaction
The neutralization reaction between HNO₃ (a strong acid) and OH⁻ (a strong base) can be represented as:
\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \]

### Step 2: Identify the Limiting Reactant
In this case, we have:
- 0.5 moles of HNO₃ (which provides 0.5 moles of H⁺)
- 0.3 moles of OH⁻

Since the reaction requires one mole of OH⁻ for each mole of H⁺, we can see that OH⁻ is the limiting reactant because we have less of it (0.3 moles) compared to H⁺ (0.5 moles).

### Step 3: Use the Enthalpy of Neutralization
The enthalpy of neutralization is given as -57.1 kJ per mole of water formed. Since only 0.3 moles of OH⁻ are available, this means that only 0.3 moles of water can be formed.

### Step 4: Calculate the Heat Released
To find the total heat released, we multiply the number of moles of the limiting reactant (OH⁻) by the enthalpy of neutralization:
\[ \text{Heat released} = \text{moles of OH}^- \times \text{enthalpy of neutralization} \]
\[ \text{Heat released} = 0.3 \, \text{mol} \times (-57.1 \, \text{kJ/mol}) = -17.13 \, \text{kJ} \]

### Step 5: Final Answer
The heat released when mixing the solutions is approximately **17.1 kJ** (the negative sign indicates that the reaction is exothermic).